Solved Assignments

CVEN 444 Assignment 4 due 6/16/03 The assignment will be review problems and will not be covered in class. You will need to show free-body diagrams, use an engineering format and be neat!

Problem 1

For the beam cross-section shown, determine

whether the failure of the beam will be initiated by

crushing of concrete or yielding of steel.

a) fc = 3500 psi As = 9 in2

b) fc =7500 psi As = 4.5 in2

Also determine whether the section satisfies ACI

Code requirement

a)

Assume that fy = 60 ksi, fc =3.5 ksi, As =9 in2 1= 0.85

Solve using equilibrium

( )( )( )( )

2

s y

c

9 in 60 ksi

0.85 0.85 3.5 ksi 14 in.

12.97 in.

C T

A fa

f b

=

= =

=

Compute c

1

12.97 in.15.25 in.

0.85

ac

= = =

Check the c/d ratio

15.25 in.0.598 0.6

25.5 in.

c

d= = >/

therefore, the beam is transition zone, the steel will yield

before the concrete crushes..

Check the min of the beam

( )( )

2

s 9.0 in 0.025214 in. 25.5 in.

A

bd = = =

Compute the minimum

y

min

c

y

200 2000.00333

60000 0.00333

3 3 35000.00296

60000

f

f

f

= =

= = =

0.0252 > 0.00333 satisfies the ACI code.

b)

Assume fy = 60 ksi, fc =7.5 ksi, As =4.5 in2

Compute the 1 value.

c1

40000.85 0.05

1000

7500 40000.85 0.05 0.675

1000

f

=

= =


( )( )( )( )

2

s y

c

4.5 in 60 ksi

0.85 0.85 7.5 ksi 14 in.

3.03 in.

C T

A fa

f b

=

= =

=

Compute c

1

3.03 in.4.48 in.

0.675

ac

= = =

Check the c/d ratio

4.48 in.0.176 0.375

25.5 in.

c

d= = 0.00433 satisfies the ACI code.

Problem 2

Calculate the nominal moment strength of the beam

a) fc = 4500 psi b) fc =7000 psi

Also determine whether the section satisfies ACI Code requirement.

a) Singly reinforced beam fy = 60 ksi, fc =4.5 ksi, As =4(1.0 in2 ) =4.0 in

2

c1

40000.85 0.05

1000

4500 40000.85 0.05 0.825

1000

f

=

= =


( )( )( )( )

2

s y

c

4 in 60 ksi

0.85 0.85 4.5 ksi 12 in.

5.23 in.

C T

A fa

f b

=

= =

=

Compute c

1

5.23 in.6.34 in.

0.825

ac

= = =

Check the c/d ratio

6.34 in.0.309 0.375

20.5 in.

c


Compute the nominal moment

( )( )

( )

n s y

2

u n

2

5.23 in.4 in 60 ksi 20.5 in. 4292.4 k-in

2

0.9 4292.4 k-in 3863.16 k-in

aM A f d

M M

=

= =

=

= =

b) Singly reinforced beam fy = 60 ksi, fc =7 ksi, As =4(1.0 in2 ) =4.0 in

2

Compute the 1 value.

c1

40000.85 0.05

1000

7000 40000.85 0.05 0.70

1000

f

=

= =


( )( )( )( )

2

s y

c

4 in 60 ksi

0.85 0.85 7 ksi 12 in.

3.36 in.

C T

A fa

f b

=

= =

=

Compute c

1

3.36 in.4.80 in.

0.7

ac

= = =

Check the c/d ratio

4.80 in.0.234 0.375

20.5 in.

c



( )( )

( )

n s y

2

u n

2

3.36 in.4 in 60 ksi 20.5 in. 4516.8 k-in

2

0.9 4516.8 k-in 4065.12 k-in

aM A f d

M M

=

= =

=

= =

b)For the l-beam use fy = 60 ksi, fc =4.5 ksi, As =6(0.79 in2 )= 4.74 in

2 and 1 = 0.825.

Assume that the 12-in section is going to work, however if it does not need to use area

concrete = 4-in.*12-in. +16-in.*x Solve using equilibrium

( )( )( )( )

2

s y

c

4.74 in 60 ksi

0.85 0.85 4.5 ksi 12 in.

6.20 in.

C T

A fa

f b

=

= =

=

Therefore, the uniform distributed load extends beyond the 4 in segment, so rework the

problem.

( )( ) ( )( )

( )( )( )

( )( ) ( ) ( )( )( )( )

s y c c

s y c

c

2

c

0.85 4 in. 12 in. 0.85 16 in.

0.85 4 in. 12 in.

0.85 16 in.

4.74 in 60 ksi 0.85 4.5 ksi 4 in. 12 in.

0.85 4.5 ksi 16 in.

1.65 in.

T C

A f f f x

A f fx

f

=

= +

=

=

=

Compute c

1

4.0 in. 1.65 in. 5.65 in.

5.65 in.6.85 in.

0.825

a

ac

= + =

= = =

Check the c/d ratio

6.85 in.0.304 0.375

22.5 in.

c

d= = 0.00335 so it satisfies the ACI standards.

Compute the nominal moment can be done by either solving for location of the center of

the compression zone to compute the moment or break the moment into two sections one

for the (4 in.)(12 in.) area and the other for (1.65 in.) (16 in.) area.

( )( )( )

( )( )( )

1 2n c c1 c c2

u n

0.85 0.852 2

4.0 in.0.85 4.5 ksi 4 in. 12 in. 22.5 in.

2

1.65 in.0.85 4.5 ksi 1.65 in. 16 in. 22.5 in. 4.0 in.

2

3763.8 k-in. 1784.8 k-in.

5548.6 k-in.

a aM f A d f A d

M M

= +

=

+ +

= +

=

=

= ( )0.9 5548.6 k-in. 4993.8 k-in=

b)For an L-beam use fy = 60 ksi, fc =7 ksi, As =6(0.79 in2 )= 4.74 in

2 and 1 =0.7.

Assume that the 12-in section is going to work, however if it does not need to use area

concrete = 4-in.*12-in. +16-in.*x Solve using equilibrium

( )( )( )( )

2

s y

c

4.74 in 60 ksi

0.85 0.85 7 ksi 12 in.

3.98 in.

C T

A fa

f b

=

= =

=

Compute c

1

3.98 in.5.69 in.

0.7

ac

= = = a 0.00418 so it satisfies the ACI standards.


( )( )

( )

n s y

2

u n

2

3.98 in.4.74 in 60 ksi 22.5 in. 5833.0 k-in.

2

0.9 5833.0 k-in. 5249.7 k-in

aM A f d

M M

=

= =

=

= =

Problem 3

Calculate the safe distributed load intensity that the beam can carry. Assume that the

only dead load is the weight of the beam (use 150 lb/ ft3). Solve for a distributed load for

the live load.

u DL LL1.2 1.6w w w= +

Use

fc = 4000 psi

fy =60000 psi

fy = 60 ksi, fc =4 ksi, As =4(1.0 in2 ) =4.0 in

2 1 =0.85


( )( )( )( )

2

s y

c

4 in 60 ksi

0.85 0.85 4 ksi 12 in.

5.88 in.

C T

A fa

f b

=

= =

=

Compute c

1

5.88 in.6.92 in.

0.85

ac

= = =

Check the c/d ratio

6.92 in.0.308 0.375

22.5 in.

c



( )( )

( )

n s y

2

u n

2

5.88 in.4 in 60 ksi 22.5 in. 4694.4 k-in

2

0.9 4694.4 k-in 4224.96 k-in

aM A f d

M M

=

= =

=

= =

Compute the maximum moment.

( )

2

uu 2

2

8

8

8 4224.96 k-in0.4075 k/in. 4.89 k/ft

12 in24 ft

1 ft

MwlM w

l

w w

= =

= = =

Assume that the load is a live load and the dead load is

( )( )2

3 1 ft0.15 k/ft 12 in. 25 in.12 in.

0.3125 k/ft.

DL

=

=

Compute live load

( )

u DLu DL LL LL

LL

1.21.2 1.6

1.6

4.89 k/ft. 1.2 0.3125 k/ft2.822 k/ft

1.6

w ww w w w

w

= + =

= =

CVEN 444 Assignment 5 due 6/18/03 The assignment will be review problems and will not be covered in class. You will need to show free-body diagrams, use an engineering format and be neat!

Compute the stresses in the compression steel, fs, for the cross-sections. Compute the

nominal moment strength and determine the reduction factor for determining the ultimate

moment.

fc = 5000 psi

fy =60000 psi

a)

fy = 60 ksi, fc =5 ksi, As =3(1.0 in2 ) =3.0 in

2 As =2(0.31 in

2 ) =0.62 in

2

c1

40000.85 0.05

1000

5000 40000.85 0.05 0.80

1000

f

=

= =

Assume the compression steel yields

( ) ( )( )( )( )

2 2s s y

c

3 in 0.62 in 60 ksi

0.85 0.85 5 ksi 10 in.

3.36 in.

A A fa

f b

= =

=

Compute c

1

3.36 in.4.2 in.

0.8

ac

= = =

Check the strain

s cu

4.2 in. 2.5 in.0.003 0.00121

4.2 in.

c d

c

= = =

Check the yield strain of steel is

y

y

s

60 ksi0.00207

29000 ksi

f

E = = =

Therefore 0.00121 is not greater than 0.00207 so the compression steel does not yield.

Use the equation for finding the c either by iterations or use

( )

( )

s y s s cu c 1

s s cu s y2 s s cu

c 1 c 1

E 0.85

E E0

0.85 0.85

c dA f A f b c

c

A A f A dc c

f b f b

= +

+ =

Plug in to the equation and solve the quadratic equation.

( )

( )( ) ( )( )( )( )( ) ( )

( )( )( )( )( ) ( )

s s cu s y2 s s cu

c 1 c 1

2 2

2

2

2

E E0

0.85 0.85

0.62 in 87 ksi 3.00 in 60 ksi0

0.85 5 ksi 10 in. 0.8

0.62 in 87 ksi 2.5 in.

0.85 5 ksi 10 in. 0.8

0 3.70765 3.96618

A A f A dc c

f b f b

c c

c c

= +

= +

=

Solve using a quadratic equation

( ) ( )

2

2

4

2

3.70765 3.70765 4 3.96618

2

4.575 in.

b b acc

a

c

=

=

=

Check the stress fs

( )

s s s cu

4.575 in. 2.5 in.29000 ksi 0.003

4.575 in.

39.46 ksi

c df E

c

= =

=

=


( )( ) ( )( )

s seff

2 23 in 0.62 in0.0136

10 in. 17.5 in. 10 in. 17.5 in.

A A

bd bd

= =

= =

Compute the minimum

y

min

c

y

200 2000.00333

60000 0.00354

3 3 50000.00354

60000

f

f

f

= =

= = =


Determine the c/d ratio

4.575 in.0.2614 0.375

17.5 in.

c

d= = <

therefore, the beam is in the tension region

and = 0.9


( ) ( )

( )( ) ( )( )( ) ( )

( )( )( )

( )

n s y s s s s

2 2

2

u n

2

0.8 4.575 in.3 in 60 ksi 0.62 in 39.46 ksi 17.5 in.

2

0.62 in 39.46 ksi 17.5 in. 2.5 in.

2804.21 k-in.

0.9 2804.21 k-in. 2523.8 k-in

aM A f A f d A f d d

M M

= +

=

+

=

=

= =

b) For the doubly reinforced beam use fy = 60 ksi, fc =5 ksi, As =4(1.0 in2 ) =4.0 in

2 As

=2(0.79 in2 ) =1.58 in

2 and 1=0.8 Assume the compression steel yields

( ) ( )( )( )( )

2 2s s y

c

4 in 1.58 in 60 ksi

0.85 0.85 5 ksi 12 in.

2.85 in.

A A fa

f b

= =

=

Compute c

1

2.85 in.3.56 in.

0.8

ac

= = =

Check the strain

s cu

3.56 in. 2.5 in.0.003 0.00089

3.56 in.

c d

c

= = =

Therefore 0.00089 is not greater than 0.00207 so the compression steel does not yield.


( )

( )( ) ( )( )( )( ) ( )( )

( )( )( )( )( ) ( )

s s cu s y2 s s cu

c 1 c 1

2 2

2

2

2

E E0

0.85 0.85

1.58 in 87 ksi 4.00 in 60 ksi0

0.85 5 ksi 12 in. 0.8

1.58 in 87 ksi 2.5 in.

0.85 5 ksi 12 in. 0.8

0 2.51324 8.42279

A A f A dc c

f b f b

c c

c c

= +

= +

=


( ) ( )

2

2

4

2

2.51324 2.51324 4 8.42279

2

4.419 in.

b b acc

a

c

=

=

=

Check the stress fs

( )

s s s cu

4.419 in. 2.5 in.29000 ksi 0.003

4.419 in.

37.78 ksi

c df E

c

= =

=

=


( )( ) ( )( )

s seff

2 24 in 1.58 in0.00938

12 in. 21.5 in. 12 in. 21.5 in.

A A

bd bd

= =

= =

Compute the minimum

y

min

c

y

200 2000.00333

60000 0.00354

3 3 50000.00354

60000

f

f

f

= =

= = =


Determine the c/d ratio

4.419 in.0.2055 0.375

21.5 in.

c

d= = <

therefore, the beam is in the tension region

and = 0.9


( ) ( )

( )( ) ( )( )( ) ( )

( )( )( )

( )

n s y s s s s

2 2

2

u n

2

0.8 4.419 in.4 in 60 ksi 1.58 in 37.78 ksi 21.5 in.

2

1.58 in 37.78 ksi 21.5 in. 2.5 in.

4692.06 k-in.

0.9 4692.06 k-in. 4222.9 k-in


M M

= +

=

+

=

=

= =

CVEN 444 Assignment 6 due 6/20/03 The assignment will be review problems and will not be covered in class. Use an engineering format and be neat! For the given sections (a),(b) and (c) assume fc = 4000 psi and fy =60000 psi

1. At failure determine whether the precast section will act similar to a rectangular section or as flanged section and check to see if the tension steel has yielded

2. Check whether the section satisfy ACI requirement. 3. Calculate the nominal moment of the section

a) For the T-Beam, use fy = 60 ksi, fc =4 ksi, As =6(1.0 in2 ) =6.0 in

2 1 =0.85

Assume a rectangular section so that b= 30 in.

( )( )( )( )

2

s y

c

6 in 60 ksi

0.85 0.85 4 ksi 30 in.

3.53 in.

A fa

f b= =

=

The a value is greater than 3 in. therefore the beam must be analyzed as T-beam.

( )

( )

( )( ) ( )( )( )( )( )

s y c eff w f c w

s y c eff w f

c

2

0.85 0.85

0.85

0.85

6 in 60 ksi 0.85 4 ksi 30 in. 12 in. 3 in.

0.85 4 ksi 12 in.

4.324 in.

T C

A f f b b h f b a

A f f b b ha

f b

=

= +

=

=

=

Compute c

1

4.324 in.5.09 in.

0.85

ac

= = =

Compute the c/d ratio

5.09 in.0.283 0.375

18 in.

c

d= = tension controlled = 0.9


( )( )

s

26 in0.0111

18 in. 30 in.

A

bd =

= =

Compute the minimum

y

min

c

y

200 2000.00333

60000 0.00333

3 3 40000.00316

60000

f

f

f

= =

= = =



( )

( )( )( )

( )( )( )

fn c eff w f c w

u n

0.85 0.852 2

3.0 in.0.85 4.0 ksi 30 in. 12 in. 3 in. 18 in.

2

4.324 in.0.85 4.0 ksi 12 in. 4.324 in. 18 in.

2

3029.4 k-in. 2794.1 k-in.

5823.5 k-in.

0.9 5

h aM f b b h d f b a d

M M

= +

=

+

= +

=

=

= ( )823.5 k-in. 5241.2 k-in.=

Compute the L-Beam using fy = 60 ksi, fc =4 ksi, As =5(1.0 in2 ) =5.0 in

2 1 =0.85


( )( )( )( )

2

s y

c

5 in 60 ksi

0.85 0.85 4 ksi 28 in.

3.15 in.

A fa

f b= =

=

The a value is greater than 3 in. therefore the beam must be analyzed as L-beam.

( )

( )

( )( ) ( )( )( )( )( )

s y c eff w f c w

s y c eff w f

c

2

0.85 0.85

0.85

0.85

5 in 60 ksi 0.85 4 ksi 28 in. 15 in. 3 in.

0.85 4 ksi 15 in.

3.282 in.

T C

A f f b b h f b a

A f f b b ha

f b

=

= +

=

=

=

Compute c

1

3.282 in.3.86 in.

0.85

ac

= = =


3.86 in.0.143 0.375

27 in.

c



( )( )

s

25 in0.0066

27 in. 28 in.

A

bd =

= =



( )

( )( )( )

( )( )( )

fn c eff w f c w

u n

0.85 0.852 2

3.0 in.0.85 4.0 ksi 28 in. 15 in. 3 in. 27 in.

2

3.282 in.0.85 4.0 ksi 15 in. 3.282 in. 27 in.

2

3381.3 k-in. 4244.6 k-in.

7625.9 k-in.

0.9 7

h aM f b b h d f b a d

M M

= +

=

+

= +

=

=

= ( )625.9 k-in. 6863.3 k-in.=

Compute the L-Beam using fy = 60 ksi, fc =4 ksi, As =4(0.79 in2 ) =3.16 in

2 1 =0.85


( )( )( )( )

2

s y

c

3.16 in 60 ksi

0.85 0.85 4 ksi 18 in.

3.10 in.

A fa

f b= =

=

The a value is less than 3.5 in. therefore the beam must be analyzed as rectangular beam.

Compute c

1

3.10 in.3.64 in.

0.85

ac

= = =


3.64 in.0.202 0.375

18 in.

c



( )( )

s

23.16 in0.00975

18 in. 18 in.

A

bd =

= =


Treat it as a rectangular section and compute the nominal moment

( )

( )( )

( )

n s y

2

u n

2

3.10 in.3.16 in 60 ksi 18 in.

2

3118.9 k-in.

0.9 3118.9 k-in. 2807.0 k-in.

aM A f d

M M

=

=

=

=

= =

CVEN 444 Assignment 7 due 6/23/03 The assignment will be review problems and will not be covered in class. Use an engineering format and be neat!

1) What is the minimum thickness of a ribbed one-way slab, if one end is continuous and length is 18 ft?

From table 9.5 a from ACI for a continuous slab

12 in.18 ft

1 ft11.68 in. Use 12.0 in.

18.5 18.5

l

= =

2) What is the pattern loading to obtain the maximum moment at 1 (center of the beam) for uniform dead load, wD, and a uniform live load, wL. Use the first figure to draw

the qualitative influence lines and the second figure to show the loading pattern.

3) What is the pattern loading to obtain the maximum shear at 2 (center of the beam) for uniform dead load, wD, and a uniform live load, wL. Use the first figure to draw the

qualitative influence lines and the second figure to show the loading pattern.

4. Determine beff , if bw =10 in., L1 = 7 ft, and L2 = 24 ft. and hf = 4 in. for both the spandrel beam(L-beam and T-beam)

For the spandrel beam (From 8.10.3 ACI code)

( )

eff w

f w

actual w

12 in.24 ft

1 ft 10 in. 34 in.

12 12

6 6 4 in. 10 in. 34 in.

12 in.7 ft

clear distance 1 ft 0.5* 10 in. 52 in.

to next web 2

Lb b

h b

b b

+ = + =

+ = + =

= + = + =

Use 34 in.

For the T beam

( )

eff

f w

actual w

12 in.24 ft

1 ft 72 in.

4 4

16 16 4 in. 10 in. 74 in.

12 in.7 ft

clear distance 1 ft 0.5* 10 in. 52 in.

to next web 2

Lb

h b

b b

= =

+ = + =

= + = + =

Use 52 in.

5. Design the eight-span east west in figure. A typical 1-ft wide design strip is shaded. A partial section through this strip is shown. The beams are assumed to

be 14 in. wide. The concrete strength is 4500 psi and the reinforcement strength

is 60 ksi. The live load is 140 psf and dead load of 80 psf.

Compute the 1 value for the concrete

1

4500 40000.85 0.05 0.825

1000

= =

Design for 1-ft wide strip, as if it were a singly reinforced section From table 9.5a

12 in.18 ft

1 ft.9.0 in.

24 24

l

= = for an external bay

12 in.18 ft

1 ft.7.7 in.

28 28

l

= = for an internal bay

Use a h = 9 in. and assume that d ~8 in. (0.75-in cover and use a #4 bar d =0.5-in.)

Compute self-weight of the beam

( ) ( ) ( )3self1 ft. 1 ft.

12 in. 9 in. 0.15 k/ft12 in. 12 in.

0.1125 k/ft

w bh

= =

=

Compute the moment action on the beam or 1 ft strip.

( ) ( )

( )( )

u DL LL

22

uu

1.2 1.6 1.2 0.080 k/ft 0.1125 k/ft 1.6 0.140 k/ft

0.455 k/ft

0.455 k/ft 18 ft.

10 10

14.742 k-ft. 176.9 k-in.

w w w

w lM

= + = + +

=

= =

=

Assume the steel is tension controlled.

nn

176.9 k-in.

0.9

196.56 k-in.

MM

= =

=

Assume that d-a/2 ~0.9d =0.9(8 in.) = 7.2 in. Solve for As

( )( )

nn s y s

y

s

2

2

2

196.56 k-in.

60 ksi 7.2 in.

0.455 in

MaM A f d A

af d

A

= =

=

=

Use 0.455 in2 per 1 ft strip. Check the minimum amount of steel

y

min

c

y

200 2000.00333

60000 0.003354

3 3 45000.003354

60000

f

f

f

= =

= = =

The area of steel min*bd=0.003354*(12-in)(8-in) = 0.322 in2 therefore we will need to

use 0.455 in2 per 1 ft strip. Use #5 bar, which will result in a d=9 in-(0.75 in cover +

0.625 in/2) =7.93 in. The spacing between bars is

( )

( )

s b b

s

2

2

12 in.12 in. s

0.31 in12 in.

0.455 in

8.17 in. Use s= 8 in.

A A As

A

s

= =

=

=

Compute a with As = 0.31 in2 * (12 in./8 in.) = 0.465 in

2 per foot

( )( )( )( )

2

s y

c

0.465 in 60 ksi

0.85 0.85 4.5 ksi 12 in.

0.608 in.

A fa

f b= =

=

Compute c

1

0.608 in.0.74 in.

0.825

ac

= = =

The c/d = 0.74 in/7.93 in = 0.093 < 0.375, therefore tension controlled and = 0.9 Compute Mn for the strip.

( )

( )( )

n s y

2

2

0.608 in.0.465 in 60 ksi 7.93 in.

2

212.77 k-in. 196.56 k-in. OK!

aM A f d

=

=

=

Check the thermal and shrinkage reinforcement requirement These bars are

perpendicular to the reinforcements for the one-way slab development.

( )( ) 2s 0.0018 0.0018 12 in. 9 in. 0.1944 inA bh= = =

Use 0.194 in2 per 1 ft strip. The maximum allowable spacing is

( )max

5 5 9 in. 45 in. 18 in.

18 in.

hs

= ==

Compute the spacing for the thermal bars using #4 bars

( )2

2

0.2 in12 in.

0.194 in

12.35 in. Use s= 12.0 in.

s

=

=

Use #4 bar at 12.0 in spacing.


1. For the beam create a envelope for the shear and for the bending moment. If the loads wL = 3.0 k/ft and wD =1.2 k/ft

You can compute the shear and bending moment diagrams using superposition tables, Visual analysis to find the reactions, There were six loadings, I looked at to obtain the envelope. Uniform load (dead and live) over the entire beam: Ray = 46.38k Rby=156.42 k, Rcy=156.42k and Rdy= 46.38k

Loading

0

1

2

3

4

5

6

7

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x (ft)

kip

s

Shear

-100

-80

-60

-40

-20

0

20

40

60

80

100

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x (ft)

kip

s

Bending Moment

-400

-300

-200

-100

0

100

200

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x (ft)

Mo

me

nt

(k-f

t)

Uniform load (dead and live) over 2/3 of beam: Ray = 44.77k Rby=163.94 k, Rcy=96.74k and Rdy= 4.155k

Load

0

1

2

3

4

5

6

7

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x

kip

s

Shear and Bending Moment Diagrams

-400

-300

-200

-100

0

100

200

300

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x (ft)

k &

k-f

t

Shear Bending Moment

Uniform load (dead and live) over 2/3 of beam: Ray =4.155 k Rby=96.74k, Rcy=163.94 k and Rdy= 44.77k

Load

0

1

2

3

4

5

6

7

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x(ft)

kip

s

load

Shear and Bending Moment Diagrams

-400

-300

-200

-100

0

100

200

300

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x (ft)

k &

k-f

t


Uniform load (dead and live) split over the two outside sections of the beam: Ray =54.53k Rby=88.27k, Rcy=88.27 k and Rdy= 54.53k

load

0

1

2

3

4

5

6

7

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x (ft)

kip

s

load

Shear and Bending Moment Diagram

-200

-150

-100

-50

0

50

100

150

200

250

300

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x (ft)

k &

k-f

t


Uniform load (dead and live) split over the center sections of the beam: Ray =2.55k Rby=104.25k, Rcy=104.25 k and Rdy= 2.55k

load

0

1

2

3

4

5

6

7

0 5 10 15 20 25 30 35 40 45 50 55 60 65

X (ft)

kip

s

load

Shear and Bending Moment Diagram

-300

-200

-100

0

100

200

300

0 5 10 15 20 25 30 35 40 45 50 55 60

x (ft)

k &

k-f

t


The shear envelope is defined as

Possible Shear

-150

-100

-50

0

50

100

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x (ft)

kip

s

DL only DL+LL DL+2/3LL DL+2/3LL(2)

Split Middle Max Shear Min Shear

Shear Envelope

-150

-100

-50

0

50

100

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x(ft)

kip

s

Maximum Shear Minimum Shear

The moment envelope is defined as

All Possible Bending Moments

-400

-300

-200

-100

0

100

200

300

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x (ft)

k-f

t

DL only DL+LL DL+2/3LL DL+2/3LL(2)

Split Middle Maximum Minimum

Bending Moment Envelope

-400

-300

-200

-100

0

100

200

300

0 5 10 15 20 25 30 35 40 45 50 55 60 65

x (ft)

k-f

t

Maximum Moment Minimum Moment

5.5

Design the simply supported beam as rectangular sections.

Given

fc = 5000 psi normal weight concrete

fy =60000 psi

a) Compute the moment of the beam

( ) ( )

( )( )

u DL LL

22

uu

1.2 1.6 1.2 0.50 k/ft 1.6 1.000 k/ft

2.20 k/ft

2.20 k/ft 20 ft.

8 8

110 k-ft. 1320 k-in.

w w w

w lM

= + = +

=

= =

=

Compute the 1 value

1

5000 40000.85 0.05

1000

0.8

=

=

Assume a single layer of steel, so d = h. - 2.5 in.. Assume that the tension controlled so

that = 0.9 and Mu = 110 k-ft Assume a k=0.30 and 1= 0.80 k = 1*k = 0.80*(0.30)=0.24

un

1320 k-in

0.9

1466.67 k-in

MM

= =

=

Find a from k = 0.24 and compute the Ru , needed for a singly reinforced beam.

( )( )( )

u c

net

0.85 12

0.240.85 5 ksi 0.24 1

2

0.8976 ksi

kR f k

A

=

=

=

Compute the size of the beam and assume b~0.55d, you can use any reasonable b

2 n

u

3 3 3

3

3

1466.67 k-in.

0.8976 ksi

1634.0 in 0.55 1634.0 in

1634.0 in14.37 in.

0.55

14.5 in.

0.55 7.9 in. 8.0 in.

Mbd

R

d

d

d

b d

= =

= =

= =

=

=

Compute the total area of steel needed as if it were a singly reinforced beam.

( )

( )

( )( )( )( )( )

c 1

s total

y

2

0.3 14.5 in. 4.35 in.

0.85 5 ksi 8 in. 0.8 4.35 in.0.85

60 ksi

1.92 in

c kd

f b cA

f

= = =

= =

=

Select the steel so that tension steel is greater or equal to 1.92 in2

Select 2 # 9 bars (1.00 in2) for the tension steel (AT = 2.00 in

2)

Compute the actual a

( )( )( )( )

2

s y

c

2.00 in 60 ksi

0.85 0.85 5 ksi 8 in.

3.53 in.

A fa

f b= =

=

Compute c

1

3.53 in.4.41 in.

0.8

ac

= = =

The c/d = 4.41 in/14.5 in = 0.304 < 0.375, therefore tension controlled and = 0.9 Compute Mn.

( )

( )( )

n s y

2

2

3.53 in.2.0 in 60 ksi 14.5 in.

2

1528.24 k-in.

aM A f d

=

=

=

The Mu for the beam and error is

( )n n 0.9 1528.24 k-in.

1375.4 k-in. 114.6 k-ft

114.6 k-ft 110 k-ftoverdesigned 100 % 4.2 %

110 k-ft

M M= =

=

= =


( )( )

s

22 in0.01724

8 in. 14.5 in.

A

bd =

= =

Compute the minimum

y

min

c

y

200 2000.00333

60000 0.00354

3 3 50000.00354

60000

f

f

f

= =

= = =


The beam is over-designed by 4.2. %, which is less than 10%. Check to see if the bars

will fit in the beam using 2 bars (#9 with #3 stirrups)

( )

( ) ( )

b stirrup2 cover

1

8 in. 2 1.128 in. 2 1.5 in. 0.375 in.

1

1.994 in.

b nd ds

n

+=

+=

=

and spacing will be 1.994 in. between bars.

b) Compute the moment of the beam. Assume that the beam has h=20 in. and b= 10 in. so that the self-weight of the beam is

( ) ( ) ( )3self1 ft. 1 ft.

20 in. 10 in. 0.15 k/ft12 in. 12 in.

0.208 k/ft

w bd

= =

=

Compute the factored loads

( ) ( )

( )

u DL LL

L

1.2 1.6 1.2 0.208 k/ft 1.6 0.0 k/ft

0.25 k/ft

1.6 10 kips 16 kips

w w w

P

= + = +

=

= =

The shear and bending moment diagrams, the reactions are 10.5 kips and Mmax =92.5 k-ft

Bending Moment

0

20

40

60

80

100

0 5 10 15 20

ft

k-f

t

Shear

-15

-10

-5

0

5

10

15

0 5 10 15 20

ft

kip

s

Compute the 1 value

1

5000 40000.85 0.05

1000

0.8

=

=


that = 0.9 and Mu = 92.5 k-ft Assume a k=0.25 and 1= 0.80 k = 1*k = 0.80*(0.25)=0.2

un

92.5 k-ft

0.9

102.78 k-ft 1233.33 k-in.

MM

= =

=


( )( )( )

u c

net

0.85 12

0.20.85 5 ksi 0.2 1

2

0.765 ksi

kR f k

A

=

=

=


2 n

u

3 3 3

3

3

1233.33 k-in.

0.765 ksi

1612.2 in 0.55 1612.2 in

1612.2 in14.31 in.

0.55

14.5 in.

0.55 7.87 in. 8.0 in.

Mbd

R

d

d

d

b d

= =

= =

= =

=

=

Go back and recomputed the moment

( ) ( ) ( )3self1 ft. 1 ft.

17 in. 8 in. 0.15 k/ft12 in. 12 in.

0.142 k/ft

w bh

= =

=


( ) ( )u DL LL1.2 1.6 1.2 0.142 k/ft 1.6 0.0 k/ft

0.17 k/ft

w w w= + = +

=

The maximum factored moment will be 88.5 k-ft

un

88.5 k-ft

0.9

98.33 k-ft 1180 k-in.

MM

= =

=

Compute d

2 n

u

3 3 3

3

3

1180 k-in.

0.765 ksi

1542.5 in 0.55 1542.5 in

1542.5 in14.10 in.

0.55

14.5 in.

0.55 7.76 in. 8.0 in.

Mbd

R

d

d

d

b d

= =

= =

= =

=

=

Compute the total area of steel needed

( )

( )

( )( )( )( )( )

c 1

s total

y

2

0.25 14.5 in. 3.625 in.

0.85 5 ksi 8 in. 0.8 3.625 in.0.85

60 ksi

1.64 in

c kd

f b cA

f

= = =

= =

=

Select the steel so that tension steel is around 1.64 in2


2)


( )( )( )( )

2

s y

c

1.58 in 60 ksi

0.85 0.85 5 ksi 8 in.

2.79 in.

A fa

f b= =

=

Compute c

1

2.79 in.3.49 in.

0.8

ac

= = =


( )

( )( )

n s y

2

2

2.79 in.1.58 in 60 ksi 14.5 in.

2

1242.44 k-in.

aM A f d

=

=

=


( )n n 0.9 1242.44 k-in.

1118.19 k-in. 93.2 k-ft

93.2 k-ft 88.5 k-ftoverdesigned 100 % 5.3 %

88.5 k-ft

M M= =

=

= =


( )( )

s

21.53 in0.01362

8 in. 14.5 in.

A

bd =

= =

Compute the minimum

y

min

c

y

200 2000.00333

60000 0.00354

3 3 50000.00354

60000

f

f

f

= =

= = =


The beam is over-designed by 5.9. %, which is less than 10%. Check to see if the bars

will fit in the beam using 2 bars (#8 with #3 stirrups)

( )

( ) ( )

b stirrup2 cover

1

8 in. 2 1.00 in. 2 1.5 in. 0.375 in.

1

2.25 in.

b nd ds

n

+=

+=

=


c) Compute the moment of the beam. Assume that the beam has h=20 in. and b= 10 in. so that the self-weight of the beam is

( ) ( ) ( )3self1 ft. 1 ft.

20 in. 10 in. 0.15 k/ft12 in. 12 in.

0.208 k/ft

w bd

= =

=


( ) ( )

( )

u DL LL

L

1.2 1.6 1.2 0.208 k/ft 1.6 0.0 k/ft

0.25 k/ft

1.6 5 kips 8 kips

w w w

P

= + = +

=

= =

The shear and bending moment diagrams, the reactions are 10.5 kips and Mmax =52.5 k-ft

Bending Moment

0

10

20

30

40

50

60

0 5 10 15 20

ft

k-f

t

Shear

-15

-10

-5

0

5

10

15

0 5 10 15 20

ft

kip

s

Compute the 1 value

1

5000 40000.85 0.05

1000

0.8

=

=


that = 0.9 and Mu = 52.5 k-ft Assume a k=0.25 and 1= 0.80 k = 1*k = 0.80*(0.25)=0.2

un

52.5 k-ft

0.9

58.33 k-ft 700 k-in.

MM

= =

=


( )( )( )

u c

net

0.85 12

0.20.85 5 ksi 0.2 1

2

0.765 ksi

kR f k

A

=

=

=


2 n

u

3 3 3

3

3

700 k-in.

0.765 ksi

915.0 in 0.55 915.0 in

915.0 in11.85 in.

0.55

12.5 in.

0.55 6.52 in. 7.0 in.

Mbd

R

d

d

d

b d

= =

= =

= =

=

=

Go back and recomputed the moment

( ) ( ) ( )3self1 ft. 1 ft.

15 in. 7 in. 0.15 k/ft12 in. 12 in.

0.109 k/ft

w bh

= =

=


( ) ( )u DL LL1.2 1.6 1.2 0.142 k/ft 1.6 0.0 k/ft

0.17 k/ft

w w w= + = +

=

The maximum factored moment will be 46.5625 k-ft

un

46.5625 k-ft

0.9

51.74 k-ft 620.83 k-in.

MM

= =

=

Compute d

2 n

u

3 3 3

3

3

620.83 k-in.

0.765 ksi

811.6 in 0.55 811.6 in

811.6 in11.38 in.

0.55

11.5 in.

0.55 6.26 in. 7.0 in.

Mbd

R

d

d

d

b d

= =

= =

= =

=

=

Use d= 12.5 in and b= 7 in to save on re-computing the moment again. Compute the total

area of steel needed

( )

( )

( )( )( )( )( )

c 1

s total

y

2

0.25 12.5 in. 3.125 in.

0.85 5 ksi 7 in. 0.8 3.125 in.0.85

60 ksi

1.24 in

c kd

f b cA

f

= = =

= =

=

Select the steel so that tension steel is around 1.24 in2


2)


( )( )( )( )

2

s y

c

1.2 in 60 ksi

0.85 0.85 5 ksi 7 in.

2.42 in.

A fa

f b= =

=

Compute c

1

2.42 in.3.03 in.

0.8

ac

= = =


( )

( )( )

n s y

2

2

2.42 in.1.2 in 60 ksi 12.5 in.

2

812.87 k-in.

aM A f d

=

=

=


( )n n 0.9 812.87 k-in.

731.59 k-in. 60.97 k-ft


46.56 k-ft

M M= =

=

= =


( )( )

s

21.2 in0.01371

7 in. 12.5 in.

A

bd =

= =

Compute the minimum

y

min

c

y

200 2000.00333

60000 0.00354

3 3 50000.00354

60000

f

f

f

= =

= = =


The beam is over-designed by 30.3. %, which is greater than 10%. If we were to use 3#5

bars we would get 3.9% over-design. Check to see if the bars will fit in the beam using 2

bars (#7 with #3 stirrups)

( )

( ) ( )

b stirrup2 cover

1

7 in. 2 0.875 in. 2 1.5 in. 0.375 in.

1

1.5 in.

b nd ds

n

+=

+=

=



1. Design a beam to handle Mu= 350 k-ft fc = 5 ksi and fy =60 ksi.

Assume that the weight of the beam has been included in the ultimate

moment. Check that the beam has met the ACI standard. Sketch the

final design. Use b= 0.5d. You may want to use a double layer of

steel.

Compute the 1 value

1

5000 40000.85 0.05

1000

0.8

=

=

Assume that the tension controlled so that = 0.9 Assume a k=0.275 and

1= 0.8, because fc =5 ksi so that k = 1*k = 0.8*(0.275) =0.22

( )( )( )

u c

'0.85 ' 1

2

0.220.85 5 ksi 0.22 1

2

0.83215 ksi

kR f k

=

=

=

You can use various k value to design the beam as long as k < 0.375 Find

bd2 from b= 0.5 d

u

2 2uu

u

2

3

350 k-ft 12 in.

0.9 1 ft.

0.83215 ksi

5608 in

M

MR bd bd

R

bd

= =

=

=

Assume that b= 0.5 d and solve for d

3 3

3

3

0.5 5608 in

5608 in22.38 in.

0.5

d

d

=

= =

and

b= 0.5(22.38 in.) = 11.19 in.

Use d =22.5 in and b = 12.in.

The dimensions and maximum moment have been determined. Assume that

k = 0.22 and a = kd= 0.22*(22.5in.) = 4.95 in. and compute the amount of

steel needed.

( )( )( )( )

cy s c s

y

s

2

0.850.85

0.85 5 ksi 12 in. 4.95 in.

60 ksi

4.21 in

f baf A f ba A

f

A

= =

=

=

Select 4 # 9 bars, so that As = 4(1.00 in2) = 4.00 in

2. Compute the new a and

c values.

( )( )( )( )

2

y s

c

1

60 ksi 4.00 in

0.85 0.85 5 ksi 12 in.

4.706 in.

4.706 in.5.882 in.

0.8

f Aa

f b

ac

= =

=

= = =

Check the min and the beam will meet the condition ACI 10.5.1

y

min min

c

y

200 2000.00333

60000 larger of 0.00354

3 3 50000.00354

60000

f

f

f

= =

= =

= =

Check the value.

( )( )

2

s 4.00 in

12 in. 22.5 in.

0.0148 0.00354 OK!

A

bd = =

=

Check to see if the steel is in the tension controlled.

t

5.882 in.0.261 0.375 ok!

22.5 in.

or

22.5 in. 5.882 in.0.003 0.003

5.882 in.

0.008475 0.005 OK!

c

d

d c

c

= =

= =

=

Compute the Mn for the beam

( )( )2n y s4.706 in.

60 ksi 4.00 in 22.5 in.2 2

4835.3 k-in.

aM f A d

= =

=


( )n n 0.9 4835.3 k-in.

4351.8 k-in. 362.65 k-ft


350 k-ft

M M= =

=

= =

The beam is over-designed by 3.61 %, which is less than 10%. Check to see

if the bars will fit in the beam using 4 bars (#9 with #3 stirrups)

( )

( ) ( )

b stirrup2 cover

1

12 in. 4 1.128 in. 2 1.5 in. 0.375 in.

3

1.25 in.

b nd ds

n

+=

+=

=

and spacing will be 1.25 in. between bars. If you use 7 #7 bars (4.20 in2) ,

which would result in 8.16% over-designed. However, you would need to

use a double layer of steel.

2. Design an interior span for a continuous beam, which will cover 8 bays (or spans). The beam has a clear distance of 26 ft and fc = 4 ksi

and fy =60 ksi. Design the beam for the positive and negative

moments with a SDL = 1.4 k/ft and LL= 2.5 k/ft. Check that the

beam has met the ACI standard. Sketch the final designs (You will

need to have two sketches, one for the positive moment and one for

the negative moment).

First you will need to compute the size of the beam. For a rough estimate

use 1-in per foot of clear span or h=26 in. and b = 0.55(26 in. 3.5 in.) =

12.375 in. Use b = 13 in.

( ) ( ) ( )

self-weight

31 ft. 1 ft.13 in. 26 in. 0.150 k/ft12 in. 12 in.

0.352 k/ft

w bh=

=

=

Compute the factored load of the beam.

( )( ) ( )( )u DL LL1.2 1.6

1.2 0.352 k/ft 1.4 k/ft 1.6 2.5 k/ft

6.1 k/ft

w w w= +

= + +

=

The load for a continuous beam with (> 2 spans) and design for the interior

negative moment of the beam and 2.5 k/ft/1.75 k/ft < 3 (ACI 8.3.3). The

moment will have the largest magnitude so it will determine the dimensions

of the beam.

( )( )22

u nu

6.1 k/ft 26 ft.

11 11

375.0 k-ft

w lM = =

=

Assume that the tension controlled so that = 0.9 Assume a k=0.3 and 1=

0.85, because fc =4 ksi so that k = 1*k = 0.85*(0.300) =0.255

( )( )( )

u c

'0.85 ' 1

2

0.2550.85 4 ksi 0.255 1

2

0.7565 ksi

kR f k

=

=

=

You can use various k value to design the beam as long as k < 0.375

Find bd2 from b= 0.55 d

u

2 2uu

u

2

3

375.0 k-ft 12 in.

0.9 1 ft.

0.7565 ksi

6610 in

M

MR bd bd

R

bd

= =

=

=


3 3

3

3

0.55 6610 in

6610 in22.91 in.

0.55

d

d

=

= =

and

b= 0.55(22.91 in.) = 12.6 in.

Use d =23.5 in and b = 13.in. Now go back and solve with the new area and

use a double layer of steel so h = 27 in. and dead load will be w = 0.366 k/ft

so that wu = 6.12 k/ft and Mu = 376.1 k-ft and using same assumptions, d =

22.93-in. and b=12.61-in. So stick with d= 23.5-in. and 13-in.


k = 0.255 and a = kd= 0.255*(23.5in.) = 5.99 in. and compute the amount

of steel needed.

( )( )( )( )

cy s c s

y

s

2

0.850.85

0.85 4 ksi 13 in. 5.99 in.

60 ksi

4.41 in

f baf A f ba A

f

A

= =

=

=

Select 4#8 bars and 2#7 bars, so that As = 4(0.79 in2)+2(0.60 in

2) = 4.36 in

2.

Compute the new a and c values.

( )( )( )( )

2

y s

c

1

60 ksi 4.36 in

0.85 0.85 4 ksi 13 in.

5.919 in.

5.919 in.6.963 in.

0.85

f Aa

f b

ac

= =

=

= = =


y

min min

c

y

200 2000.00333

60000 larger of 0.00333

3 3 40000.00316

60000

f

f

f

= =

= =

= =

Check the value.

( )( )

2

s 4.36 in

13 in. 23.5 in.

0.0143 0.00333 OK!

A

bd = =

=


t

6.963 in.0.296 0.375 ok!

23.5 in.

or

23.5 in. 6.963 in.0.003 0.003

6.963 in.

0.00713 0.005 OK!

c

d

d c

c

= =

= =

=


( )( )2n y s5.919 in.

60 ksi 4.36 in 23.5 in.2 2

5373.5 k-in.

aM f A d

= =

=


( )n n 0.9 5373.5 k-in.

4836.1 k-in. 403.0 k-ft


376.1 k-ft

M M= =

=

= =


if the bars will fit in the beam using 4 bars (#8 with #3 stirrups) per row

( )

( ) ( )

b stirrup2 cover

1

13 in. 4 1.0 in. 2 1.5 in. 0.375 in.

3

1.75 in.

b nd ds

n

+=

+=

=


Compute the factored load of the beam for the positive moment

( )( ) ( )( )u DL LL1.2 1.6

1.2 0.366 k/ft 1.4 k/ft 1.6 2.5 k/ft

6.12 k/ft

w w w= +

= + +

=

The load for a continuous beam with (> 2 spans) and design for the interior

positive moment of the beam and 2.5 k/ft/1.77 k/ft < 3 (ACI 8.3.3). The

moment will have the largest magnitude so it will determine the dimensions

of the beam.

( )( )22

u nu

6.12 k/ft 26 ft.

16 16

258.6 k-ft

w lM = =

=

Determine the Ru from the equation and find a

( )( )

u

2 uu u 2

u 2

258.6 k-ft 12 in.

0.9 1 ft.

13.0 in. 24.5 in.

0.4419 ksi

M

MR bd R

bd

R

= =

=

=

Compute k from

( )( )

2 uu c

c

u

c

2'0.85 ' 1 2 0

2 0.85

2 0.4419 ksi21 1 1 1

0.85 0.85 4 ksi

0.140

RkR f k k k

f

Rk

f

= + =

= =

=

The value for a = k*d = 0.140*24.5 in. =3.43 in. The area of steel can be

computed

( )( )( )( )

cy s c s

y

s

2

0.850.85

0.85 4 ksi 13 in. 3.43 in.

60 ksi

2.52 in

f baf A f ba A

f

A

= =

=

=

Select 2#8 bars and 2#7 bars, so that As = 2(0.79 in2)+2(0.60 in

2) = 2.78 in

2.


( )( )( )( )

2

y s

c

1

60 ksi 2.78 in

0.85 0.85 4 ksi 13 in.

3.77 in.

3.77 in.4.44 in.

0.85

f Aa

f b

ac

= =

=

= = =

Check the value.

( )( )

2

s 2.78 in

13 in. 24.5 in.

0.0087 0.00333 OK!

A

bd = =

=


t

4.44 in.0.181 0.375 ok!

24.5 in.

or

24.5 in. 4.44 in.0.003 0.003

4.44 in.

0.0136 0.005 OK!

c

d

d c

c

= =

= =

=


( )( )2n y s3.77 in.

60 ksi 2.78 in 24.5 in.2 2

3771.9 k-in.

aM f A d

= =

=


( )n n 0.9 3771.9 k-in.

3394.7 k-in. 282.9 k-ft


258.6 k-ft

M M= =

=

= =


if the bars will fit in the beam using 4 bars (2#8+2#7 with #3 stirrups) per

row

( )

( ) ( ) ( )

b stirrup2 cover

1

13 in. 2 1.0 in. 2 0.875 in. 2 1.5 in. 0.375 in.

3

1.83 in.

b nd ds

n

+=

+=

=



1. Design a doubly reinforced concrete beam to handle Mu= 380 k-ft fc = 5 ksi and fy =60 ksi. Assume that the weight of the beam has been

included in the ultimate moment. Check that the beam has met the

ACI standard. Sketch the final design. Use b= 0.5d. Use a

compression steel, As = 0.4-0.5 Anet. You may want to use a double

layer of steel.

Compute the 1 value

1

5000 40000.85 0.05

1000

0.8

=

=

Assume a double layer of steel, so d = 24 in. - 3.5 in. = 20.5-in. Assume that

the tension controlled so that = 0.9 and Mu = 380 k-ft Assume a k=0.35

and 1= 0.80 k = 1*k = 0.80*(0.350) =0.28

un

380 k-ft

0.9

422.22 k-ft 5066.67 k-in

MM

= =

=

Find a from k = 0.28 and compute the Ru , needed for a singly reinforced

beam.

( )( )( )

u c0.85 12

0.280.85 5 ksi 0.28 1

2

1.0234 ksi

kR f k

=

=

=

Compute the size of the beam

2 n

u

3 3 3

3

3

5066.67 k-in.

1.0234 ksi

4950.8 in 0.5 4950.8 in

4950.8 in21.47 in.

0.5

21.5 in. 0.5 11 in.

Mbd

R

d

d

d b d

= =

= =

= =

= =


( )

( )

( )( )( )( )( )

c 1

s total

y

3

0.35 21.5 in. 7.525 in.

0.85 5 ksi 12 in. 0.8 8.575 in.0.85

60 ksi

4.69 in

c kd

f b cA

f

= = =

= =

=

Compute the Anet for the beam assume that 50% of the Anet is in the

compression.

( )

( )

2s total

net

net

2

2 2

s

4.69 in

1 % 1 0.5

3.13 in

0.5 3.13 in 1.56 in

AA

A

A

= =+ +

=

= =

Select the steel so that tension steel is greater or equal to 4.69 in2 and the

compression steel is greater than 1.56 in2 and the net steel is greater than

3.13 in2


2) and 2#8 bars

(0.79 in2) for the compression steel (As=1.58 in

2) and net steel of Anet = 3.16

in2.

Assume the compression steel yields and is at d= 2.5 in.

( ) ( )( )( )( )

2 2s s y

c

4.74 in 1.58 in 60 ksi

0.85 0.85 5 ksi 11 in.

4.06 in.

A A fa

f b

= =

=

Compute c

1

4.06 in.5.07 in.

0.8

ac

= = =

Check the strain

s cu

5.07 in. 2.5 in.0.003 0.001521

5.07 in.

c d

c

= = =

Check the yield strain of steel is

y

y

s

60 ksi0.00207

29000 ksi

f

E = = =

Therefore 0.01521 is not greater than 0.00207 so the compression steel does

not yield.

Or check the values

( ) ( )( )( )( )

1 ceff

y y

0.85 87

87

0.8 0.85 5 ksi 2.5 in. 870.021232

60 ksi 21.5 in. 87 60

f d

f d f

=

Check the values.

( )( )

( )( )

2

s

2

s

4.74 in

11 in. 21.5 in.

0.02004

1.58 in

11 in. 21.5 in.

0.00668

A

bd

A

bd

= =

=

= =

=

The eff = 0.02004 0.00668 = 0.01336 is not greater than 0.021232 so the compression steel has not yielded

Use the equation for finding the c either by iterations or use

( )

( )

s y s s cu c 1

s s cu s y2 s s cu

c 1 c 1

E 0.85

E E0

0.85 0.85

c dA f A f b c

c

A A f A dc c

f b f b

= +

+ =


( )

( )( ) ( )( )( )( )( ) ( )

( )( )( )( )( ) ( )

s s cu s y2 s s cu

c 1 c 1

2 2

2

2

2

E E0

0.85 0.85

1.58 in 87 ksi 4.74 in 60 ksi0

0.85 5 ksi 11 in. 0.8

1.58 in 87 ksi 2.5 in.

0.85 5 ksi 11 in. 0.8

0 3.92888 9.1885

A A f A dc c

f b f b

c c

c c

= +

= +

=


( ) ( )

2

2

4

2

3.92888 3.92888 4 9.1855

2

5.577 in.

b b acc

a

c

=

=

=

Check the stress fs

( )

s s s cu

5.577 in. 2.5 in.29000 ksi 0.003

5.577 in.

48.00 ksi

c df E

c

= =

=

=


y

min min

c

y

200 2000.00333

60000 larger of 0.003536

3 3 50000.003536

60000

f

f

f

= =

= =

= =

The eff = 0.01336 > 0.003536 so it is OK. Check to see if the steel is in the tension controlled.

t

5.577 in.0.259 0.375 ok!

21.5 in.

or

21.5 in. 5.577 in.0.003 0.003

5.577 in.

0.008566 0.005 OK!

c

d

d c

c

= =

= =

=

Compute the Mn , the nominal moment

( ) ( )

( )( ) ( )( )( ) ( )

( )( )( )

n s y s s s s

2 2

2

2

0.8 5.577 in.4.74 in 60 ksi 1.58 in 48.00 ksi 21.5 in.

2

1.58 in 48.00 ksi 21.5 in. 2.5 in.

5459.78 k-in


= +

=

+

=


( )n n 0.9 5459.78 k-in.

4913.8 k-in. 409.5 k-ft


380 k-ft

M M= =

=

= =

The beam is over-designed by 7.76. %, which is less than 10%. Check to see


( )

( ) ( )

b stirrup2 cover

1

11 in. 3 1.0 in. 2 1.5 in. 0.375 in.

2

2.125 in.

b nd ds

n

+=

+=

=

and spacing will be 2.125 in. between bars. There are 2 #8 bars in the

compression zone.

Design a T-beam with a length of L1= 18-ft and spacing between beam, L=8

ft. The beam must be designed to handle a positive moment of 200 k-ft and a

negative moment of 300 k-ft (assume the weight of the beam is included in

the moments) with a maximum depth of 24-in and slab thickness of 3 in. Use

fc = 5 ksi and fy =60 ksi.

Design for the larger moment, 300 k-ft and the compression zone would be

in the web of the beam, So you can design the beam like a singly reinforced

beam.

Compute the 1 value

1

5000 40000.85 0.05

1000

0.8

=

=



( )( )( )

u c

'0.85 ' 1

2

0.240.85 5 ksi 0.24 1

2

0.8976 ksi

kR f k

=

=

=


bd2 from b= 0.6 d

u

2 2uu

u

2

3

300 k-ft 12 in.

0.9 1 ft.

0.8976 ksi

4456 in

M

MR bd bd

R

bd

= =

=

=


3 3

3

3

0.6 4456 in

4456 in19.51 in.

0.6

d

d

=

= =

and

b= 0.6(19.51 in.) = 11.71 in.


Determine beff in tension and compression

The compression width is

( )eff f w eff

w

12 in.18 ft

1 ft54 in.

4 4

16 16 3 in. 12 in. 60 in. 54 in.

12 in.8 ft

clear distance 1 ft0.5* 12 in. 60 in.

to next web 2

L

b h b b

b

= =

+ = + = =

+ = + =

The tension width is

( )

( )eff eff tension

eff comp.

12 in.18 ft

1 ft21.6 in. 21 in.

10 10

b 54 in.

Lb b

= = =

=


k = 0.24 and a = kd= 0.24*(19.5in.) = 4.68 in. and compute the amount of

steel needed.

( )( )( )( )

cy s c s

y

s

2

0.850.85

0.85 5 ksi 12 in. 4.68 in.

60 ksi

3.98 in

f baf A f ba A

f

A

= =

=

=

Select 4 # 7 bars and 4 #6 bars, so that As = 4(0.6 in2)+4(0.44 in

2) = 4.16 in

2.


( )( )( )( )

2

y s

c

1

60 ksi 4.16 in

0.85 0.85 5 ksi 12 in.

4.894 in.

4.894 in.6.118 in.

0.8

f Aa

f b

ac

= =

=

= = =

Check the minimum area, min, and the beam will meet the condition ACI 10.5.1

( )( )

( )( )

( )( )

c 2

w

y

2

effminy

c 2

eff

y

2

min

6 6 500012 in. 19.5 in. 1.65 in

60000

200 20054 in. 19.5 in. 3.51 insmallest of

60000larger of

3 3 500054 in. 19.5 in. 3.72 in

60000

1.65 in

fb d

f

b dAf

fb d

f

A

= =

= == = =

=

Check the value.

( )( )

2

s 4.00 in

12 in. 22.5 in.

0.0148 0.00354 OK!

A

bd = =

=


t

6.118 in.0.314 0.375 ok!

19.5 in.

or

19.5 in. 6.118 in.0.003 0.003

6.118 in.

0.006562 0.005 OK!

c

d

d c

c

= =

= =

=

Compute the Mn for the T-beam

( )( )2n y s4.894 in.

60 ksi 4.16 in 19.5 in.2 2

4256.4 k-in.

aM f A d

= =

=


( )n n 0.9 4256.4 k-in.

3830.8 k-in. 319.23 k-ft


300 k-ft

M M= =

=

= =

The beam is over-designed by 6.41 %, which is less than 10%. The bars

(4#7 and 4#6 bars) will fit into the 21 in.(beff in tension). The #7 bars would

fit into the 12-in

Compute the maximum capacity of the T-beam, where the flange is in

compression Mn. The value for d will be 21in- 2.5 in =18.5 in.

( )( )( )fn c eff f3 in.

0.85 0.85 5 ksi 54 in. 3 in. 18.5 in.2 2

11704.5 k-in. 975.4 k-ft.

hM f b h d

= =

=

So the T-beam can be designed as a single reinforced concrete beam. The

minimum amount steel required is

( )( )

( )( )

2

eff

y

min

c 2

eff

y

2

min

200 20054 in. 19.5 in. 3.51 in

60000larger of

3 3 500054 in. 19.5 in. 3.72 in

60000

3.72 in

b df

Af

b df

A

= =

=

= =

=

Use 5 #8 bars that As = 5(0.79 in2) = 3.95 in

2.

( )( )( )( )

2

y s

c

1

60 ksi 3.95 in

0.85 0.85 5 ksi 54 in.

1.033 in.

1.033 in.1.291 in.

0.8

f Aa

f b

ac

= =

=

= = =

Using a double layer of steel for 5 bars in a12 in flange the d=21-in 3.5 in.=

17.5 in instead of 19.5 in.

( )( )2n y s1.033 in.

60 ksi 3.95 in 17.5 in.2 2

4025.1 k-in.

aM f A d

= =

=


( )n n 0.9 4025.1 k-in.

3622.6 k-in. 301.88 k-ft


200 k-ft

M M= =

=

= =

To match the minimum amount of steel will result in over-designing the

beam by 50%. The double layer of 3 #8 bars would fit into the 12-in

( )

( ) ( )

b stirrup2 cover

1

12 in. 3 1.0 in. 2 1.5 in. 0.375 in.

2

2.625 in.

b nd ds

n

+=

+=

=



1. Calculate the basic development lengths in tension for the following deformed bars embedded in normal weight concrete.

(a) For #6 and #9 bar. Given fc = 6 ksi and fy =60 ksi. (b) For #14 and #18 bar. Given fc = 5 ksi and fy =60 ksi and fy =75 ksi.

Assume clear spacing =2db, clear cover = 1.5-in. on each side and bars

are not spliced.

(a) #6 bar, bar placement =1.0, coating, =1.0, size of the bar,

=0.8, and concrete, = 1.0. and db =0.75 in. and clear space between bars is 2(0.75 in) =1.5 in.

b

b

clear 0.75 in.1.5 in. 1.875 in.

space 2 21.125 in.

1.5 in. 1.5 in. 0.75 in.1.125 in.

2 2

d

c cd

+ = + =

= =+ + = =

The bar is assumed be under no tension so Ktr =0.

tr

b

1.125 in. 0.01.5 1.5 Use 1.5

0.75 in.

c K

d

+ += =

Check the concrete fc

c 100

6000 77.5 100 OK.

f

=

Compute the development length

( )( )( )( )( )

( )

yd

b c tr

b

d b

3

40

1.0 1.0 0.8 1.03 6000030.98

40 1.56000

33.94 30.98 0.75 in.

23.3 in.

fl

d f c K

d

l d

= +

= =

= =

=

or you can use 24 in. from table 10.2

(b) #9 bar, bar placement =1.0, coating, =1.0, size of the bar,

=1.0, and concrete, = 1.0. and db =1.128 in. and clear space between bars is 2(1.128 in) =2.256 in.

b

b

clear 1.128 in.1.5 in. 2.06 in.

space 2 21.692 in.

2.256 in. 2.256 in. 1.128 in.1.692 in.

2 2

d

c cd

+ = + =

= =+ + = =

The bar is assumed be under no tension so Ktr =0.

tr

b

1.692 in. 0.01.5 1.5 Use 1.5

1.128 in.

c K

d

+ += =

Check the concrete fc

c 100

6000 77.5 100 OK.

f

=

Compute the development length

( )( )( )( )( )

( )

yd

b c tr

b

d b

3

40

1.0 1.0 1.0 1.03 6000038.73

40 1.56000

38.73 38.73 1.128 in.

43.69 in.

fl

d f c K

d

l d

= +

= =

= =

=

or you can use 44 in.

(c) #14 bar, bar placement = 1.0, coating, = 1.0, size of the

bar, = 1.0, and concrete, = 1.0.and db =1.693 in.

( )( )( )( )

( )

yd

b c

d b

20

1.0 1.0 1.0 1.06000042.43

205000

42.43 42.43 1.693 in.

71.83 in.

fl

d f

l d

=

= =

= =

=

Use ld = 72 in.

#18 bar, bar placement = 1.0, coating, = 1.0, size of the bar,

= 1.0, and concrete, = 1.0. and db =2.257 in.

( )( )( )( )

( )

yd

b c

d b

20

1.0 1.0 1.0 1.07500053.03

205000

53.03 53.03 2.257 in.

119.70 in.

fl

d f

l d

=

= =

= =

=

Use ld = 120 in.

2. Calculate the basic embedment lengths, if the bars are used as compression reinforcement and the concrete is sand-lightweight.

(a) For #6 and #9 bar. Given fc = 6 ksi and fy =60 ksi. (b) For #14 and #18 bar. Given fc = 5 ksi and fy =60 ksi and fy =75 ksi.

(a) #6 bar, concrete, = 1.3. and db =0.75 in. Compute the development length

( )

( )

y

db d

c

b b

y

db b

600000.02 0.02 15.49

186000

0.0003 0.0003 60000 18.0

18 18 0.75 in.

13.5 in.

fl l

fd d

f

l d

= = = =

= =

= =

=

The final development length is

( )d db

1.3 13.5 in.

17.55 in. Use 18.0 in.

l l=

=

=

#8 bar, concrete, = 1.3. and db =1.00 in. Compute the development length

( )

( )

y

db d

c

b b

y

db b

600000.02 0.02 15.49

186000

0.0003 0.0003 60000 18.0

18 18 1.128 in.

20.304 in.

fl l

fd d

f

l d

= = = =

= =

= =

=


( )d db

1.3 20.304 in.

26.40 in. Use 27.0 in.

l l=

=

=

(b) #14 bar, , = 1.3.and db =1.693 in. . Compute the development length

( )

( )

y

db d

c

b b

y

db b

600000.02 0.02 16.97

18.005000

0.0003 0.0003 60000 18.0

18.0 18.0 1.693 in.

30.474 in.

fl l

fd d

f

l d

= = = =

= =

= =

=


( )d db

1.3 30.474 in.

39.62 in. Use 40.0 in.

l l=

=

=

#18 bar, concrete, = 1.3. and db =2.257 in. Compute the development length

( )

( )

y

db d

c

b b

y

db b

750000.02 0.02 21.22

22.505000

0.0003 0.0003 80000 22.5

22.50 22.50 2.257 in.

50.78 in.

fl l

fd d

f

l d

= = = =

= =

= =

=


( )d db

1.3 50.78 in.

66.02 in. Use 67.0 in.

l l=

=

=


1. Determine the development length required for the bars shown .fc =4-

ksi and fy = 60-ksi. Check the anchorage in the column. If it is not

satisfactory, design an anchorage using a 180o hook and check

adequacy.

Determine the development length for the bars.The bar coefficients are

(bar placement) = 1.3, (coating)= 1.0, and (light weight concrete) = 1.0 and #10 bar is db = 1.27 in..

( )( )( )

( )

yd

b c

d b

d

20

1.3 1.0 1.0 6000061.66

20 4000

61.66

61.66 1.27 in. 78.3 in. Use 79 in.

fl

d f

l d

l

=

= =

=

= =

The total length available in the column is l = 51 in. 1.5 in. = 49.5 in.,

therefore there is insufficient length for start development. We do not know

the transverse bars, spacing or strength of the bars. However,(c+Ktr)/db can

be no greater than 2.5. So use 2.5 and check and =1.0 (bar size)

( )( )( )( )

( )

( )

yd

b trc

b

d b

d

3

40

3 1.3 1.0 1.0 1.0 6000037.0

40 4000 2.5

37.0

37.0 1.27 in. 46.98 in. Use 47 in.

fl

d c Kf

d

l d

l

=

+

= =

=

= =

So you could argue that there is sufficient space if the transverse loading is

enough to have, (c+Ktr)/db > 2.5. However, a hook would be a better method

to ensure that there is sufficient development length. The development

length is computed by

( )

hd

b c

hd b

hd

1200

120018.97

4000

18.97

18.97 1.27 in. 24.1 in. Use 25 in.

l

d f

l d

l

=

= =

=

= =

There is no factor used in hooks and the other reduction factors are for

cover, yield strength of the steel fy = 60 ksi , = 1, concrete =1, and

excessive reinforcement is unknown so use = 1. and ties assume =1

So there is sufficient length for a hook

2. Problem 10.5 from the text --An 18-ft normal weight concrete cantilever beam is subject to factored Mu =3500 k-in and a factored

shear Vu = 32.4 k at the face of the support. Design the top

reinforcement and the appropriate embedment of 90o hook into the

concrete wall to sustain the external shear and moment. Given fc =

4.5 ksi and fy =60 ksi.

Compute the 1 value

1

4500 40000.85 0.05

1000

0.825

=

=

Assume a single layer of steel, so d = h. - 2.5 in.. Assume that the tension

controlled so that = 0.9 and Mu = 3500 k-in Assume a k=0.30 and 1=

0.825, k = 1*k = 0.825*(0.30)=0.2475

un

3500 k-in

0.9

3888.89 k-in

MM

= =

=

Find a from k = 0.2475 and compute the Ru , needed for a singly reinforced

beam.

( )( )( )

u c

net

0.85 12

0.24750.85 4.5 ksi 0.2475 1

2

0.8295 ksi

kR f k

A

=

=

=

Compute the size of the beam and assume b~0.55d, you can use any

reasonable b

2 n

u

3 3 3

3

3

3888.89 k-in.

0.8295 ksi

4688.0 in 0.55 4688.0 in

4688.0 in20.43 in.

0.55

20.5 in.

0.55 11.3 in. 12.0 in.

Mbd

R

d

d

d

b d

= =

= =

= =

=

=


( )

( )

( )( )( )( )( )

c 1

s total

y

2

0.3 20.5 in. 6.15 in.

0.85 4.5 ksi 12 in. 0.825 6.15 in.0.85

60 ksi

3.88 in

c kd

f b cA

f

= = =

= =

=

Select the steel so that tension steel is greater or equal to 3.88 in2


2)


( )( )( )( )

2

s y

c

4.00 in 60 ksi

0.85 0.85 4.5 ksi 12 in.

5.23 in.

A fa

f b= =

=

Compute c

1

5.23 in.6.34 in.

0.825

ac

= = =

The c/d = 6.34 in/20.5 in = 0.309 < 0.375, therefore tension controlled and = 0.9

Compute Mn.

( )

( )( )

n s y

2

2

5.23 in.4.0 in 60 ksi 20.5 in.

2

4292.4 k-in.

aM A f d

=

=

=


( )n n 0.9 4292.4 k-in.

3863.2 k-in.

3863.2 k-in 3500 k-inoverdesigned 100 % 10.3 %

3500 k-in

M M= =

=

= =


( )( )

s

24 in0.01626

12 in. 20.5 in.

A

bd =

= =

Compute the minimum

y

min

c

y

200 2000.00333

60000 0.003354

3 3 45000.003354

60000

f

f

f

= =

= = =


The beam is over-designed by 10.3. %, which is less than 10%. Check to see

if the bars will fit in the beam using 2 bars (#9 with #3 stirrups)

( )

( ) ( )

b stirrup2 cover

1

12 in. 4 1.128 in. 2 1.5 in. 0.375 in.

3

1.246 in.

b nd ds

n

+=

+=

=


Check the embedment length of 90o hook. The basic development length

( )

hd

b c

hd b

1200 120017.89

4500

17.89 17.89 1.128 in.

20.18 in

l

d f

l d

= = =

= =

=

Normal weight concrete, l = 1.0 and the area ratio

2

sd 2

s

Required 3.88 in0.97

Provided 4.00 in

A

A = = =

The final length

( )dh d hb 0.97 20.18 in.

19.58 in.

l l= =

=

Use ldh = 20 in. Check to see it is greater 8db=8(1.128in.) = 9.024in. or 6

in, which is correct. The depth of the 12db = 12(1.128in.) = 13.536in. use

14in.


1. For a 28-ft simply supported beam (b=16 in. and d=26.5 in) with 8#8 bars (2 layers) and fc =4-ksi and fy = 60-ksi, carrying uniform loads of

wD =1.5 k/ft and wL=3.0 k/ft. Draw the moment capacity envelope if

total of 4-bars are terminated (2 at a time) and calculate theoretical

cutoff locations, actual cutoff locations, development length, moment

capacity of a bar and verify the beam will satisfy the design and meet

ACI standards.

Compute the loading on the beam

( ) ( )u DL LL1.2 1.6

1.2 1.5 k/ft 1.6 3.0 k/ft

6.6 k/ft

w w w= +

= +

=

The load for a simply supported beam and design for the positive moment of

the beam and 4.8 k/ft

( )( )22

u nu

6.6 k/ft 28 ft.

8 8

646.8 k-ft

w lM = =

=

Assume that the tension controlled so that = 0.9 and 1= 0.85, because fc =4 ksi



c values.

( )( )( )( )

2

y s

c

1

60 ksi 6.32 in

0.85 0.85 4 ksi 16 in.

6.971 in.

6.971 in.8.201 in.

0.85

f Aa

f b

ac

= =

=

= = =


y

min min

c

y

200 2000.00333

60000 larger of 0.00333

3 3 40000.00316

60000

f

f

f

= =

= =

= =

Check the value.

( )( )

2

s 6.32 in

16 in. 26.5 in.

0.0149 0.00333 OK!

A

bd = =

=


t

8.201 in.0.309 0.375 ok!

26.5 in.

or

26.5 in. 8.201 in.0.003 0.003

8.201 in.

0.00669 0.005 OK!

c

d

d c

c

= =

= =

=


( )( )2n y s6.971 in.

60 ksi 6.32 in 26.5 in.2 2

8727.1 k-in.

aM f A d

= =

=


( )n n 0.9 8727.1 k-in.

7854.4 k-in. 654.5 k-ft


646.8 k-ft

M M= =

=

= =

The beam is over-designed by 1.2 %, which is less than 10%. Check to see if

the bars will fit in the beam using 4 bars (#8 with #4 stirrups) per row

( )

( ) ( )

b stirrup2 cover

1

16 in. 4 1.0 in. 2 1.5 in. 0.5 in.

3

2.67 in.

b nd ds

n

+=

+=

=

and spacing will be 2.67 in. between bars

For both cutoff and splicing on needs to know the development length of the

bars.



( )( )( )

( )

yd

b c

d b

d

20

1.0 1.0 1.0 6000047.43

20 4000

47.43

47.43 1.00 in. 47.43 in. Use 48 in.

fl

d f

l d

l

=

= =

=

= =

Determine the capacity of 1 bar

( )( )( )

ubar s y

2

2

6.971 in.0.9 0.79 in 60 ksi 26.5 in.

2

981.8 k-in. 81.82 k-ft.

aM A f d

=

=

=

So the Mu (1 bar) =81.82k-ft, Mu (2 bars) =163.6 k-ft, Mu (3 bars) =245.4 k-

ft, and Mu (4 bars) =327.3 k-ft. Mu (5 bar) =409.1k-ft, Mu (6 bars) =490.9 k-

ft, Mu (7 bars) =572.7 k-ft, and Mu (8 bars) =654.5 k-ft

Moment Diagram for Half Beam

0

100

200

300

400

500

600

700

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

x (ft)

Mo

men

t (k

-ft)

Bending Moment 8 Bars 6 Bars 4 Bars

Determine the development length and la for the bars. The extension length

la is either 12db=12(1.0 in.) = 12.0 in. or d= 26.5 in. Use la = 26.5 in or 27 in.

Determine where the bar cutoff points meet the moment diagram.. Obtain

the equation for the moment

( ) ( )( )

2

u 92.4 k 6.6 k/ft2

xM x x=

Set the moment equal to 490.9 k-ft and 327.3 k-ft

( )( )

2

2

490.9 k-ft. 92.4 k 6.6 k-ft2

28 148.76 0 7.13 ft. or 85.52 in. 85 in.

xx

x x

=

+ =

( )( )

2

2

327.3 k-ft. 92.4 k 6.6 k-ft2

28 98.182 0 4.16 ft. or 49.92 in. 49 in.

xx

x x

=

+ =

The location of the cut off for a two bars are 85 in.-27 in.= 58 in. (4-ft 10 in.)

and will be symmetrical about the centerline. The location for the bars to be

fully developed is 58 in + 48 in. =106 in.(8-ft 10 in.).

The location of the cut off for a 4 bar is 49 in.- 27 in.= 22 in. (1-ft 10 in.) and

The location for the bar to be fully developed is 22 in + 48 in. =70 in.(5-ft 10

in)

2. Design the compression lap splice for a column section 16-in. x 16-in. reinforce with 8 #10 bars equally spaced around all faces

a. Given fc = 5-ksi and fy =60-ksi b. Given fc = 7-ksi and fy =80-ksi

(a) #10 bar, bar placement =1.0, coating, =1.0, size of the bar,

=1.0, and concrete, = 1.0. and db =1.27 in.

( )

( )

y

db d

c

b b

y

db b

600000.02 0.02 16.97

185000

0.003 0.003 60000 18.0

18 18 1.27 in.

22.86 in.

fl l

fd d

f

l d

= = = =

= =

= =

=

For fy =60 ksi and check to see that it is greater than 12-in.

( )( )d y b0.0005

0.0005 60000 psi 1.27 in.

38.1 in. Use 39.0 in.

l f d=

=

=

(b) #10 bar, bar placement =1.0, coating, =1.0, size of the bar,

=1.0, and concrete, = 1.0. and db =1.27 in.

( )

( )

y

db d

c

b b

y

db b

800000.02 0.02 19.12

24.07000

0.003 0.003 80000 24.0

24.0 24.0 1.27 in.

30.48 in.

fl l

fd d

f

l d

= = = =

= =

= =

=

For fy =80 ksi and check to see that it is greater than 12-in.

( )( )( )( )

d y b0.0009 24

0.0009 80000 psi 24 1.27 in.

60.96 in. Use 61.0 in.

l f d=

=

=

A simply supported uniform loaded beam carries a total factored load of

4.8 k/ft (this includes the beam weight) on a clear span of 34-ft, fc =3-

ksi and fy = 40-ksi. Assume that the supports are 12-in wide and assume

the bars are available in 30-ft lengths

c. Design a rectangular beam (tension steel only) d. Determine bar cutoffs. e. Locate splices and determine lap length.

The load for a simply supported beam and design for the positive moment of

the beam and 4.8 k/ft

( )( )22

u nu

4.8 k/ft 34 ft.

8 8

693.6 k-ft

w lM = =

=



( )( )( )

u c

'0.85 ' 1

2

0.21250.85 3 ksi 0.2125 1

2

0.4843 ksi

kR f k

=

=

=


bd2 from b= 0.55 d

u

2 2uu

u

2

3

693.6 k-ft 12 in.

0.9 1 ft.

0.4843 ksi

19096 in

M

MR bd bd

R

bd

= =

=

=


3 3

3

3

0.55 19096 in

19096 in32.62 in.

0.55

d

d

=

= =

and

b= 0.55(32.94 in.) = 17.94 in.



k = 0.2125 and a = kd= 0.2125*(32.5 in.) = 6.906 in. and compute the

amount of steel needed.

( )( )( )( )

cy s c s

y

s

2

0.850.85

0.85 3 ksi 18 in. 6.906 in.

40 ksi

7.92 in

f baf A f ba A

f

A

= =

=

=



c values.

( )( )( )( )

2

y s

c

1

40 ksi 8.00 in

0.85 0.85 3 ksi 18 in.

6.972 in.

6.972 in.8.202 in.

0.85

f Aa

f b

ac

= =

=

= = =


y

min min

c

y

200 2000.005

40000 larger of 0.005

3 3 30000.00411

40000

f

f

f

= =

= =

= =

Check the value.

( )( )

2

s 8.00 in

18 in. 32.5 in.

0.013675 0.005 OK!

A

bd = =

=


t

8.202 in.0.252 0.375 ok!

32.5 in.

or

32.5 in. 8.202 in.0.003 0.003

8.202 in.

0.00887 0.005 OK!

c

d

d c

c

= =

= =

=


( )( )2n y s6.972 in.

40 ksi 8.00 in 32.5 in.2 2

9284.5 k-in.

aM f A d

= =

=


( )n n 0.9 9284.5 k-in.

8356.1 k-in. 696.3 k-ft


693.6 k-ft

M M= =

=

= =



( )

( ) ( )

b stirrup2 cover

1

18 in. 4 1.128 in. 2 1.5 in. 0.5 in.

3

3.16 in.

b nd ds

n

+=

+=

=

and spacing will be 3.16 in. between bars. So you could have bars next to

one another for splicing.

For both cutoff and splicing on needs to know the development length of the

bars.



( )( )( )

( )

yd

b c

d b

d

20

1.0 1.0 1.0 4000036.51

20 3000

36.51

36.51 1.128 in. 41.19 in. Use 42 in.

fl

d f

l d

l

=

= =

=

= =

Determine the capacity of 1 bar

( )( )( )

ubar s y

2

2

6.972 in.0.9 1.00 in 40 ksi 32.5 in.

2

1044.5 k-in. 87.04 k-ft.

aM A f d

=

=

=

So the Mu (1 bar) =87.04k-ft, Mu (2 bars) =174.1 k-ft, Mu (3 bars) =261.2 k-

ft, and Mu (4 bars) =348.2 k-ft. Mu (5 bar) =435.2k-ft, Mu (6 bars)

=522.4 k-ft, Mu (7 bars) =609.3 k-ft, and Mu (8 bars) =696.3 k-ft

Determine the development length and la for the bars. The extension length

la is either 12db=12(1.127 in.) = 13.52 in. or d= 32.5 in. Use la = 32.5 in or

33 in.

Determine where the bar cutoff points meet the moment diagram.. Obtain

the equation for the moment

( ) ( )( )

2

u 81.6 k 4.8 k/ft2

xM x x=

Set the moment equal to 552.4 k-ft and 348.3 k-ft

( )( )

2

2

522.4 k-ft. 81.6 k 4.8 k-ft2

34 217.67 0 8.55 ft. or 102.65 in. 102 in.

25.45 ft. or 305.4 in. 306 in.

xx

x x

=

+ =

( )( )

2

2

348.3 k-ft. 81.6 k 4.8 k-ft2

34 145.125 0 5.00 ft. or 60 in. 60 in.

29.0 ft. or 348 in. 348 in.

xx

x x

=

+ =

The location of the cut off for a single bar is 102 in.- 33 in.= 69 in. (5-ft 9

in.) and 306- in +33 in. = 339 in. (28.f

Documents

Solved Assignments